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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{P_2O_5}=\dfrac{42,6}{142}=0,3\left(mol\right)\)
PTHH: P2O5 + 3H2O ---> 2H3PO4
0,3------------------------------>0,6
ddA có chứa axit photphoric H3PO4, nhận biết bằng QT, H3PO4 làm QT chuyển sang màu đỏ
\(m_{ct}=m_{H_3PO_4}=0,6.98=58,8\left(g\right)\)
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
PTHH: 3Mg + 2H3PO4 ---> Mg3(PO4)2 + 3H2
LTL: \(\dfrac{0,5}{3}< \dfrac{0,6}{2}\rightarrow\)H3PO4 dư
Theo pt: \(n_{H_2}=n_{Mg}=0,5\left(mol\right)\)
\(\rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
a)Theo pt: \(n_{Fe}=n_{H_2}=0,1mol\)
\(\Rightarrow m_{Fe}=0,1\cdot56=5,6g\)
\(\%m_{Fe}=\dfrac{5,6}{12}\cdot100\%=46,67\%\)
\(\%m_{Cu}=100\%-46,67\%=53,33\%\)
c)\(m_{Cu}=12-5,6=6,4g\Rightarrow n_{Cu}=0,1mol\)
\(BTe:n_{O_2}=n_{Fe}+n_{Cu}=0,2mol\)
\(H=80\%\Rightarrow n_{O_2}=80\%\cdot0,2=0,16mol\)
\(n_{KMnO_4}=\dfrac{1}{2}n_{O_2}=0,08mol\)
\(\Rightarrow m_{KMnO_4}=0,08\cdot158=12,64g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
b, Dung dịch Y chứa NaOH.
- Cách nhận biết: Nhỏ vài giọt dd vào quỳ tím thấy quỳ chuyển xanh.
c, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Na}=2n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2.23}{10,8}.100\%\approx42,59\%\\\%m_{Na_2O}\approx57,41\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nAl=8,1/27=0,3mol ;nH2SO4=53,9/98=0,55mol
ta có pt : 2Al+3H2SO4---->Al2(SO4)3+3H2
Trước p/u: 0,3mol 0,55mol
p/u : 0,3mol 0,45mol
Saup/u: 0mol 0,1mol 0,15mol 0,45mol
=>H2SO4 dư
mH2SO4 dư =0,1.98=9,8g
b,mAl2SO43 =0,15.294=44,1g
c, mH2=0,45.2=0,9g
V H2=0,45.22,4=10,08l
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,5}{2}\) => Fe dư, HCl hết
PTHH: Fe + 2HCl --> FeCl2 + H2
0,5----------->0,25
=> VH2 = 0,25.22,4 = 5,6 (l)
b)\(n_{Fe_3O_4}=\dfrac{13,92}{232}=0,06\left(mol\right)\)
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
Xét tỉ lệ: \(\dfrac{0,06}{1}< \dfrac{0,25}{4}\) => Fe3O4 hết, H2 dư
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
0,06-->0,24------->0,18-->0,24
=> \(\left\{{}\begin{matrix}m_{Fe}=0,18.56=10,08\left(g\right)\\m_{H_2O}=0,24.18=4,32\left(g\right)\\m_{H_2\left(dư\right)}=\left(0,25-0,24\right).2=0,02\left(g\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\frac{2,7}{27}=o,1mol\)
n HCl = o,2 mol
2 Al +6 HCl →2AlCl3 + 3H2
bđ: 0,1
đang bận !
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
Ta có: \(n_{O_2}=\dfrac{1,6}{32}=0,05\left(mol\right)\)\(\Rightarrow n_{Cu}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow\%m_{CuO}=\dfrac{0,1\cdot80}{40}\cdot100\%=20\%\)
\(\Rightarrow\%m_{Fe_2O_3}=80\%\)